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T
TECHNOLOGF'".
, .,
ON THE ASYMPTOTIC PROPERTIES OF
EUCLIDEAN DIAL-A-RIDE ROUTING
by
Mordecai Haimovich*
OR 145-85
*
October 1985
University of Chicago, Graduate School of Business
This research was supported by Grant ECS-7926625 from the Systems
Theory and Operations Research Division of the National Science
Foundation.
ABSTRACT
A conjecture by Stein [1], proposing a probabilistic limit result
for the shortest possible route of a bus that has to transfer passengers
between random locations in some region of the plane, is refuted.- The
existence of such limit result remains an open question.
1.
Introduction
Let ol, d,
planar region R.
02, d2 , 03, d3 , ...
be a sequence of points in a bounded
A problem related to the scheduling of a Dial-A-Ride
transportation system is concerned with finding the shortest route for a
bus that has to transfer passengers from o.i to di for all i
n.
Let Y
be the length of such shortest route when the capacity of the bus is
unlimited.
Let Ln be the length of the shortest path (i.e. travelling
2
salesman tour) through the first n points of the sequence ol, d,
02,
d2 ,
03,.
Considering the infinite product of Lebesgue measure on the plane or
alternatively a probability space on which ol, d,
02,
d2 ,
03,
... is a
sequence of independent uniformly distributed random points in R, it has
been shown by Beardwood Halton and Hammerslev (BHH) [2] that there exists a
constant b such that:
L
lim
n = b
a.e. (almost everywhere)
(1)
new
where a is the area of the region R, and where b has been estimated to be
approximately 0.75.
Noting that Yn > L2n it follows as has already been observed in [1]
that:
y
lim inf
n > 4b
a.e.
(2)
n~o
In order to achieve an upper limit result, Stein [1] constructs the
following heuristic "two passage" algorithm:
Algorithm A
Partition R into m subregions r,
r2 , ..., rm each of area a/m.
On
the "first passage" through the regions, pick up in each subregion r all
passengers waiting there and deliver all passengers with destinations in
r.i who were picked up in rl, r2 , ..., ri_.
Then on "second passage"
through rl, r2, ..., rm, deliver the remaining passengers.
On each
-3-
passage through a subregion, the bus uses the shortest path through the
points visited at that passage.
Using algorithm A and the BHH result (1), Stein shows that
Y
lim sup -n
n-o
b JV
-
a.e.
(3)
3
and demonstrates that the routes obtained by this algorithm are
asymptotically optimal, in a class of so called "simple tours".
He
further conjectures that they are actually asymptotically optimal in the
4
class of all possible routes and that
of
Y /vi
(a.e.).
bV
b/
is actually the limit
In the next section we refute this conjecture by
slightly improving the algorithm described above and demonstrating
that:
lim sup T
n-*m
One may still ask if
Y /n
< 4
b
a
a.e.
converges at all.
(4)
This in contrast to (1)
remains an unsettled issue.
2.
Proof of (4)
Let
0. be the set of origins in r..
Let D
in ri for passengers with origins in rl, r2,
and let D
i
..., ri
be the rest of the destinations in r..
i
A above, the points in 0. U D
be the set of destinations
(define D1 = 0)
According to algorithm
are visited during the first passage
through ri, while the points in D
1
are visited on the second passage.
It seems worthwhile, however, within the 2-passage partitioning
framework of algorithm A, to:
-4-
(a) Delay some of the deliveries to destinations in D1 to the
second passage through ri (for i = 2, 3, ..., m).
(b) Delay to the second passage some of the pickups of passengers
in ri whose delivery to destinations in ri+l, ri+ 2 ,
..., rm is delayed
according to (a) above.
It remains, however, to show that such improvements are
asymptotically significant in the sense that (3) is not tight.
We shall
bound from below the reduction of cost possible through (a) and (b)
above.
To do this we consider only a part of the worthwhile delivery
delays.
Let Db be the set of destinations in D chosen according to the
1
1
following sequential procedure:
Scan the destinations in D
according to their order.1
For each
scanned p E D2 , examine the circular neighborhood of radius h-/n
1'
(h > 0) around it. If there are in the circle destinations from Di
1
which have not been selected earlier, then select the one which is
closest to p and include it in D2.
Consider now:
Algorithm B
Same as algorithm A except that delay the delivery of points in D2
for i = 2, 3, ..., m from first passage to second passage.
1That is, the order induced by their order in the sequence ol, dl,
02, d 2 , .... Also any order which is independent of their locations will
do.
2 The
reason for that special way of selecting only one at a time is
to avoid statistical dependence between the locations of the points in
1
1
-5-
Note that according to the construction of D21' the points in D1\D2
1
1
(the destinations visited on first passage through ri ) are independently
And assuming that n >> m, such that
and identically distributed in r.
hJi/n
a/m and that the maximal length of the
is negligible compared to
boundary of a subregion is
(a7m ), we may neglect the asymptotically
diminishing "boundary effects" and say that the points in D\D2
are
distributed uniformly over r..
We may try now to establish a lower bound to the reduction in cost
in comparison to algorithm A.
We first estimate the asymptotic behavior of IDl, the number of
Consider then again the sequential construction of
delayed deliveries.
D2,
stopped at some instant, when the number of D
[y m] for some non-negative y.
have not yet been selected is j
probability that a currently scanned point of D
unselected point of D
destinations which
The
will have a yet
1
in its h7n neighborhood (neglecting the
1
inaccuracy due to the event of vanishingly small probability, that the
scanned D
point falls within less than ha7n from the boundary of r)
71h
pn(j) = 1 - (1
a
/an)j
n
=
n
h2) m n/m
1 - (1n
a/m
Note that as n t
, (1
-
(where
Consequently for any
monotone convergence).
such that for all n
m nh2)m t e
denotes increasing
> 0 there exists N(&)
N(£)
-7th
2
J
n/m <
Jl ( j)
1
(e-h
2
-
n/i
is:
-6-
Let
x
m
2
be the number of
D.1
points that have been scanned so far.
Now suppose the selection process is continued, scanning the next 3 [6x-m]
points of D.
Let Aj be the number of points that get selected, then
1
Aj
[6x- n] and thus
m
6
y
j < 6x.
n/m
actually selecting a point from D
[1
-
,
e, h2(y
1 - (e
h
In each stage the probability of
lies in the interval
1
It follows (omitting some technicali-
)Y].
-
ties) that:
-nh2 (y-6x) < lim inf
< lira sup y < 1 - e-h
=
=~
6x
6x =
n-om
nxm
It follows that in probability 1, as n
X
2
y
a.e.
y and x tend to satisfy the
differential equation:
d
dx
=-(1 - e-nh2y)
ID'
y(0)
Let z(x) = y(O) - y(x).
m
i
n/rn
lim n/
n/m
Then lim(1
n-
JD21j
(= lim n
_
n/m
m
) a.e. (note that
m
i- 1
a.e.) and:
m
= 1
dz
-(h2(i
dz
m
)>
dx-1- e
h2 ( 1+ 7h2(-i
1 + h2(
m
z)
z
7th2( i m1 + nh
2
z)
'
- z)
3If there happen to be less than [6x,-] unscanned points, then scan
only the existing unscanned points.
-7-
where z(O) = 0. Consequently
z(1 -
i
1
>
. - 1
nth2( i - 1
m
z(
-
1~+h 2
i - 1
))
m
(1)m (1
i-
1
,
or explicitly:
i
i
z(1
- 1h
/mm
=zl
m
lim
limn/m
n-*w
i7th 2
>
n/h
)
1 + 2h
1
2
mm
(1
i-
)
a.e.
m
.
We evaluate now a lower bound to the total decrease in route lengths
within the subregion r..
1
i
Use the abbreviations:
= z(1
i - 1), a =
m
The increase of length of second passage through r. is at most
m
1
ID2{ ·
2h
=
4g
m- 2h( n)
-
a
1·
m
The
decrease
length
of first passage through r
of
is asymptotically
The decrease of length of first passage through r
is asymptotically
(using the BHH result (1)):
b
m(1
+
n
m
)a
m
n (1 + a
in
b
-
m b
m
) m=
mm
m
4
->m
>
41 +
a - C)
1 b
22 '
The total decrease of route length in ri is,therefore,at least:
1
in
b
2h)>
(2J2 - 2h)C 2 nakJ7
2 h)l+
h2
- 2h)n
1 i
1
i
m
and thus the total decrease of route length in the whole region is at
least:
-8-
b
Nw(
Now for
i1 m i nth 2
m2(1 2
m
m
i
2h)1 + 2h m
1i
we have:
m
1 m i - 1
i=
m(1
m
m i=l
i - 1)
m
-+
fx(l - x)dx = 1
06
'
Hence the total decrease in cost is asymptotically at least
7h 2
1 b
6(2-- 2h) 1 + 2nh2 '4n
which for
h = b/6V2 (a choice approximately maximizing this expression)
is equal to
na
nb3ly~b
722(1 +
b
I.e., instead of (3) we have almost everywhere
3
Tb
bbb
lim sup Ynn < 4
+
722(1
Tn=3
7-,
~~~36
n~~~cr,
= 4J
3
(nb2/36)
b(1 -
)j
< 4
Tb)
192(1 + 36
Using the estimate b --- 0.75 we ha
nb 2 /36
2/36) =
192(1 + Tnb
1000024 = 0.00024%
,
2b
-9which is a very slight improvement to the previous bound, an improvement
too slight to justify the complication of algorithm A, but which nonetheless refutes its conjectured optimality and the resulting assertion
regarding the limit of
Y //i.
Note that while further improvements are
certain, we havelfor the sake of analytical tractability, pursued only
the improvement implied by algorithm B.
3.
References:
[1]
,[2]
Stein, D. 11., [1978].
"An Asymptotic Probabilistic Analysis of
.a Routing Problem," Math. of Oper. Res., 3, pp. 89-101.
Beardwood, J., Halton, J. H. and J. M. Hammersley, [1959].
"The Shortest Path Through Many Points," Proc. Cambridge
Philos. Soc., 55, pp. 299-327.